3
$\begingroup$

A pseudovariety $\mathbf{V}$ of groups is join prime if for any pseudovarieties $\mathbf{V}_1, \mathbf{V}_2, \ldots,\mathbf{V}_m$, the implication $$\mathbf{V} \subseteq \mathbf{V}_1 \vee \mathbf{V}_2 \vee \cdots \vee \mathbf{V}_m \quad \Longrightarrow \quad \mathbf{V} \subseteq \mathbf{V}_i$$ holds for some $i$. A finite group is join prime if it generates a join prime pseudovariety.

It is known that all groups of order up to 7 are join prime. So it is natural to ask: is the dihedral group $D_4$ of order 8 join prime?

| cite | improve this question | | | | |
$\endgroup$
4
$\begingroup$

Yes, the pseudovariety generated by $D_4$ is join prime (and the argument shows that the same is true for the pseudovariety generated by $8$-element quaternion group). The result follows from two observations:

(1) the class ${\mathbf P}$ of finite groups whose Sylow $2$-subgroups are abelian forms a pseudovariety (i.e., this class is closed under finite products, the formation of subgroups and the formation of quotients), and
(2) any pseudovariety not contained in ${\mathbf P}$ contains $D_4$ (and $Q_8$).

Assuming Items (1) and (2), and the obvious fact that $D_4\not\in {\mathbf P}$, we argue as follows: if ${\mathbf V}(D_4)\subseteq {\mathbf V}_1\vee \cdots \vee {\mathbf V}_m$, then by Item (1) there is some $i$ such that ${\mathbf V}_i\not\subseteq {\mathbf P}$. By Item (2), ${\mathbf V}_i$ contains $D_4$, so ${\mathbf V}(D_4)\subseteq {\mathbf V}_i$.

I explain how to prove Item (2). Assume ${\mathbf V}$ is a pseudovariety containing some group $G$ with a nonabelian Sylow $2$-subgroup. We may assume that $G$ is chosen with $|G|$ minimal, and that ${\mathbf V}={\mathbf V}(G)$. Necessarily $G$ is a nonabelian, subdirectly irreducible $2$-group with monolith $M = \langle z\rangle\subseteq Z(G)$ where $z^2=1$, and $G/M$ is abelian. In particular, $G$ is $2$-step nilpotent.

Since $G/M\models [x,y]\approx 1$ and $M\models x^2\approx 1$ we get that $G\models [x,y]^2\approx 1$.

It follows from commutator collection that the set of laws of any finite $2$-step nilpotent group may be axiomatized by: the group laws, the law $[[x,y],z]\approx 1$, an exponent bound $x^m\approx 1$, and an exponent bound on the commutator subgroup $[x,y]^n\approx 1$. Thus, if the exponent of our group $G$ is $2^r$, then since $G$ is nonabelian and satisfies $[x,y]^2\approx 1$ we get that ${\mathbf V}(G)$ is exactly the class of all finite, $2$-step nilpotent $2$-groups satisfying $[x,y]^2\approx 1$ and $x^{2^r}\approx 1$. If $2^r\geq 4$, this class contains $D_4$.

But we must have $2^r\geq 4$, since otherwise $2^r\mid 2$ and $G\models x^2\approx 1$. This can't happen since $G$ is nonabelian.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.